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B
−secx⋅tan7x7
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C
sec6x⋅tanx7
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D
secx⋅tanx7
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Solution
The correct option is Csec6x⋅tanx7 We know that, In=∫secnxdx=tanx⋅secn−2xn−1+n−2n−1In−2
Putting n=8, we get I8=∫sec8xdx=tanx⋅sec6x7+67I6 ⇒I8−67I6=tanx⋅sec6x7